Thank you, that makes a lot of sense.
Since it's inside the rod, the gaussian surface of the charge would enclose less than the total charge of the rod.
So wouldn't it be
E×2πrL= (πLr^2)/(εR^2)
Which would make E=(λ1×r)/(2πεR^2)
Is this correct?
Okay thank you.
Quick question as I am working on this. The charge enclosed can only be 0 if this was a conducting rod, right? I know if this was a conducting rod, the charge enclosed would be 0, making the electric field 0. But, is there any instance where a charge enclosed can be 0 in a non...
Hello, thank you for your help.
In class we did a problem where the charge enclosed varies by r and these are the steps we took. In this problem, the radius of the pipe was 10cm, but the radius of the charge was 5cm. To find the charge enclosed we compare the densities of the pipe versus the...
Homework Statement
Below is a diagram of an infinitely long non-conducting rod of radius, R, with a uniform continuous charge distribution. The uniform linear charge density of this line is lamba1. The rod is at the center of an infinitely long, conducting pipe. The linear charge density of...